Switching Q1. (1) Explain the major components of a telephone exchange and explain their limiting factors

Transcription

1 Switching Q. () Explain the major components of a telephone exchange and explain their limiting factors Basic components of a Telephone Exchange (i) Switching unit To connect an input to a output. Limiting factor is number of connections (erlang). (ii) Control unit To handle all controlled functions. Limiting factor will be number of call attempts per unit time. (iii) Peripherals units The supporting equipment. No specific limitations (iv) Software Instructions and methodology to process a call has been written. Limiting factor will be the time taken to process a call. (2) Explain the difference between Strict and Wide sense of Non Blocking. Strict Sense of Non Blocking- To achieve non-blocking without using any rule. Ex: Close Circuit Wide Sense of Non Blocking- To achieve non-blocking with using a rule. Ex: Ben s Rule (3) A T switch will connect two subscribers in two different locations as follows. D C E A B T switch C is the Master Switch, where as D and E are Remote Switches. Each remote switch will connect to Master Switch by 6 PCM systems namely P 0 to P 3. A is speaking with B on P 0 TS8, where as B is speaking to A on P 8 TS20. (i) Explain clearly how A and B is in conversation with respect to the Master Switch.

2 (ii) Assuming only these RSU are connected to this MSU, calculate the maximum traffic that this switch can handle. TS = Time Slot C. D P 0f P 6f E P 0b P 6b P 5f P 3f P 5b P 3b 2. A speaks on P 0 TS8 B speaks on P 8 TS20 3. When A speaks to be the following two actions will occur 3. P 0f TS8 P 8b TS20 A Hello 3.2 P 8f TS20 P 0b TS8 B Hello 4. What happens at C (overview) 25µs P 0f TS8 P 8b TS20 P 8f P 0b TS8 Z= 0

3 5. The above transactions will be realized in the following way There are 3 components in a T Switch. Assuming output Control T Switch all forward PCM timeslots will be stored in the Buffer memory cyclicly by the forward PCM in the Buffer Memory. The output PCM are cyclicly addressed the control memory and the contents will give the word number of the Buffer memory to be read. In a give 25us, the control memory has to be refreshed two times by the Processor. Buffer Memory Control Memory Where to read P 0fTS8 P 8bTS20 Q2. Processor i. Explain the basic elements of a fixed telephone network. The following are the basic elements of a fixed telephone network. Geographical location of the customer. 2. Local Exchange (maximum distance a local exchange will serve is around 0km) 3. Access Network (Connection between above and 2) 4. Domestic Transport Network (interconnection between all local exchanges) 5. International Gateway 6. International Transport Network (Interconnection between all international gateways of each country for each operator) ii. Basic demarcation of telephone network can be categorized as follows. Out-side plant Switching Transmission Explain the physical demarcation of each of the above and the functions that we have to achieve from the each above.

4 Out-side Plant- The demarcation will be a telephone exchange MDF (cable side) to the geographical location of the customer. Technical function that has to be achieved will be to reproduce the electrical signal generated from the customer premises telephone to the MDF (Main Distribution Frame). Switching- When customer dials a telephone number B switching will connect to the B (B can be any telephone in the world). In general switch will capture all answered calls details for the customer billing. Transmission- The demarcation will be from one switch to another switch (normally through a MDF or DDF-Digital Distribution Frame). The technical functionality to be achieved is similar to Out-side Plant. In general OSP is distributed according to the customer s location and involve in high cost in the case of Copper network. Transmission is converged from one point to another point and much technology is deployed to enhance many channels into a single bearer. Switching carry major intelligence of the networks and it provides online computers. iii. A digital telephone exchange contains a single stage T switch, which can accommodate 52 PCM systems. Assuming all the 52 PCM systems can be utilized for telephone conversation, what is the maximum traffic in Earlangs that this switch can carry. The traffic that can be carried: Number of time slots in the switching system = 52*30*2 ( PCM system is 2 PCMs) Number of Time Slots used for a conversations = 4 Number of conversations that can be carried out in the switching system = (52*30*2)/4 = 7680 Hence maximum of 7680 Erlangs can be carried out in this switch. Two RSUs connected to the exchange are connected with PCM0 to PCM5 to one RSU, while the other RSU is connected with PCM385 PCM400. Total number of PCMs to the Master switch will be 52 (assume all these PCMs are used for RSUU). Two customers connected from these two RSUU are connected via PCM TS5 and PCM386 TS0 to the main switch. If this T switch is working as an output

5 controlled T switch, explain clearly how the T switch works when these two customers are in conversation with respect to the elements of buffer memory, controller memory and microprocessor. Note: TS = Time Slot Customer A is speaking on PCM TS5 while customer B is speaking on PCM386 TS0. Hence we observe the following switching function at the master switch. P0 f TS05 Delay of 9.5µs P386 b TS0 P386 f TS0Delay of 05.5µs P0 b TS05 Timing Chart P f TS05 P 386b TS0 P 386f P b TS05 Z= 0 In order to achieve the above switching function 3 components are needed i.e. Buffer memory- To store the forward PCMs TS information Control memory- Control information to achieve the above switching function Processor- To carry out the switching operations. Output Control T switch Here the output PCMs (Backward PCMs) is cyclically addressing the Control memory and the relevant word pertaining to output TS will provide where to read from the Buffer. Storing of input PCM TS information is cyclic at the Buffer memory. Input Control T Switch Here the input PCMs (Backward PCMs) is cyclically addressing the Control memory and the relevant word pertaining to input TS will provide where to store the relevant

6 input PCM TS information (which is according to the above switching equation). Reading of output PCM TS information is cyclic at the Buffer memory. In Summary Output Control T Switch Buffer Memory Input PCM Time Slots 4 The respective Output PCM read the correct input TS Cyclically storing into the Buffer Control Memory 2 Output PCM Time Slots 3 Address rigidly Connected Addressing will be cyclically done. Contents where to read from the Buffer Input Control T Switch Buffer Memory Input TS will be stored according to the instructions given in the control memory 3 Output PCM Time Slots 4 Input PCM Time Slots Control Memory Buffer memory will be cyclically read by the output PCM TS Cyclically Addressing into the Control Contents will give where to in the Buffer memory 2 Generally Output Control T switch widely applicable due to less errors experienced. Nevertheless for short hauls Input Control T switch can be used. Where to store in the Buffer memory of the following forward PCM P0 f TS05 P386 f TS0

7 Total of 52 PCM are available to store in the Buffer memory. Hence the Buffer memory size has to be 52 *32 words of length and width will be 8 bits. To define a word in the Buffer memory we need 4 binary bits where 9 major significant bits are for the PCM number and the least 5 significant bits are for the TS number. P0 f TS05 = ,000 2 = 0025 H = 37 0 P386 f TS0 =,00,000,000 2 = 324A H = (386 0 = 92 H =,00,000) Output Control T Switch Buffer Memory W Input PCM Time Slots W37 Cyclically storing into the Buffer W2874 W Bit PCM0 TS05 PCM386 TS0 3 Control Memory 4 The respective Output PCM read the correct input TS 2 Output PCM Time Slots Address rigidly Connected Addressing will be cyclically done. Contents where to read from the Buffer In the Control memory word number 37 will be addressed by PCM0 b TS05 and the contents is to read the Buffer memory in the word number Similarly word number 2874 is addressed by the PCM386 b TS0 and the contents of the control memory is to read word number 37. The size of the control memory is 4 bit word and 6384 words. The Processor deployed to carry out the instructions of the switching equation given above should have cycle time better than (25/52)*30*2 µs.

8 Q3.. Explain the difference between Analogue & Digital switching. In the switching module one can see all the voltages with respect to time which is generated in the telephone is called analogue switch, whereas in the digital switch you can see only samples or equivalent of samples. 2. What are the basic components of a Telephone exchange explain the functions of each. Basic components of a Telephone Exchange (i) Switching unit To connect an input to a output (ii) Control unit To handle all controlled functions. (iii) Peripherals units The supporting equipment. (iv) Software Instructions and methodology to process a call has been written. 3. Design 3-stage strict sense of non-blocking 49 inputs, 49 outputs. How many middle switches you use? Prove the total number cross points are 9. 7 () 3 () () 3 7 (7) 7 3 (3 The number of middle switch required = 7 x 2 = 3 (7) 3 7 Total number of cross points; X X X x7x3 7x7x3 3x7x7 7x7x3x3 = 9 4. If you have to design 48 input 48 out 5 stage switching Network, how do you design to obtain minimum number of cross points? Calculate the number of cross points in that network.

9 7x6 () 6 7x 7x7 7 7 (2) () () x2 (6) (2) 7 6x7 (6) (3) 2 6 Number of switches from 6 i/p or i/p should access = = 2 In order to obtain minimum Number of cross points, the links from i/p small switch has to be transpose with o/p small switch which is shown in the diagram. No.of cross points = 859 = 7x3x6 + 6x2 + 7x6 + (7x7) x+ 6x7 + 3x7x6 + 2x6 Input No.of cross points = 7x3x6 + 6x2 = 68 Middle No.of cross points = 7x6 + (7x7)x + 6x7 = 623 Output No.of cross points = 3x7x6 + 2x6 = 68 Total = 859

10 Q4. a. Today the telephone exchanges use packet switching technology, named Next Generation Networks. Explain the basic advantage of utilizing NGN networks. NGN networks will have the following advantages over the normal circuit switch network. i. Normal circuit switch network when deployed for voice the transmission network will be efficient less than 50%.This was due to the fact that from A to B a bidirectional transmission 2 parts are established for a connection and the voice is such that where both parties not speak simultaneously, when one speaks other person listens resulting the other persons voice channel is not being utilized though it is established. Hence NGN networks transport part is utilized for packet switching for 00% efficiency of the transport network where there is no fixed channels being allocated for A and B. ii. Similar concept explained above exist in the digital switching network too where circuit switching is used allocating 2 clear channels in the switching network for the communication of A and B. A and B does not talk simultaneously. Hence the efficiency of the switching network too is less than 50% n circuit switching. The advancement of computer and packet switching technologies has helped to migrate a normal circuit switch to a packet switch which is of high efficiency. iii. In the circuit switching networks integration of services such as Data, IPTV etc. was difficult and have been handled by separate networks by using IP technology for both transport and switching networks in the NGN environment leads to provide unified service to the customer through the already provided access network. Hence more utilization of access network with an unified terminal equipment with high resolution screen, keyboard, speakers will be made available in future.

11 b. The following is a part of digital switching network deployed in Sri Lanka. Each Remote Subscriber Unit is connected to Master exchange by 8 PCM systems. RSU P 0 Master Exchange (MSU) Ratmalan a P 8 RSU P 7 P 5 Moratuwa Piliyandal A B i. From the diagram what is your assessment of the number of customers that can be connected to each RSU. Each RSU is connected with 8 PCM systems i.e. 240 bi-directional Time Slots for voice. Assume a customer will speak (either originate or terminate) 6 minutes in the busy hour. The traffic offered by a customer will be 6/60= 0. Erlang. 240 bi-directional circuits can carry 240 Erlangs (assuming there is no loss) customers for each RSU can be accommodated. ii. Assuming that there are no customers connected at Ratmalana Master exchange, what is the maximum traffic that the Master exchange can handle. (Assume no other connections from Master exchange to other exchanges and this will work as a closed Switching network.) The traffic that can be carried: Number of time slots in the switching system = 6*30*2 ( PCM system is 2 PCMs) Number of Time Slots used for a conversations = 4 Number of conversations that can be carried out in the switching system = (6*30*2)/4 = 240 Hence maximum of 240 Erlangs can be carried out in this switch. iii. If A customer speaking to B customer on P03 TS05 and P4TS0. Clearly explain how digital switching is achieved at Ratmalana Master exchange. At Ratmalana master switch the following switching function will be achieved.

12 P03 f TS05 Delay of 9.5µs P4 b TS0 P4 f TS0 Delay of 05.5µs P03 b TS05 In order to achieve the above switching function P03 f TS05 and P4 f TS0 has to be stored in a buffer memory. Lets analyze the buffer memory. Buffer memory size- Since there are6 forward PCMs resulting 6*32 = 52. The size of the buffer memory should be 8bit word (time slot information is 8 bits) of 52 words. Where to store in the buffer memory? If we assume buffer memory is partition to accommodate st PCM then Time Slot a word in the buffer memory can be defined (in binary) as follows X 9 X 8 X 7 X 6 X 5.. X 0 where X 5 to X 9 represent the PCM number and the rest represent the Time Slot number. Hence PCM P03 f TS05 will store in the following word 000,000 2 = 0C5 H = (4*6+5) 0 = 229 Similarly P4 f TS0 will store in the following word 00,00 2 = AA H = (*6 2 +0*6+0) 0 = 426 Buffer Memory W 0 8 Bit word W 229 P03 f TS05 information W 426 P4 f TS0 information W 52 How to read the Buffer memory? Reading the Buffer memory will be directed by the control memory. Control memory too will have 52 words since there are 52 Time Slots pertaining to backward PCMs.

13 If we assume Control memory is partition to accommodate st PCM then Time Slot a word in the Control memory can be defined (in binary) as follows X 9 X 8 X 7 X 6 X 5.. X 0 where X 5 to X 9 represent the PCM number and the rest represent the Time Slot number. Hence PCM P03 b TS05 will address the Control memory in the following word 000,000 2 = 0C5 H = (4*6+5) 0 = 229 Similarly P4 b TS0 will store in the following word 00,00 2 = AA H = (*6 2 +0*6+0) 0 = 426 The Control memory maximum size will be of 52 words where each word will be of 9 bits to indicate from which buffer memory word the output Time Slots should read. Control Memory W 0 9 Bit word W 229 Go to W 426 of Buffer and read Will address P03 b TS05 W 426 Go to W 229 of Buffer and read Will address P4 b TS0 W 52 Processor Processor will control all these functions for every 25µs reading and writing of Time Slots will be managed. Hence if there are 52 Time Slots the processor cycle time has to be better than 25/52 *2µs. The following timing chart summarizes how the above transactions are visible in the Ratmalana switch.